# understanding DDE23 function format

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alpedhuez on 1 May 2020
Edited: alpedhuez on 4 May 2020
DDE23 function has a format of
ddex1de(t,y,Z)
How should one understand Z?

Guru Mohanty on 4 May 2020
I understand you are trying to solve system of differential equation using ‘dde23’. To do this you need to do following steps.
1. Define constant delays.
In this system of equation there are two lags i.e. (t-1) and (t-0.2).
lags = [1 0.2];
2. Define Solution History.
It Defines the first solution from which the solver starts iterations.
function s = history(t)
s = ones(3,1);
end
3. Form the equation.
function dydt = ddex1de(t,y,Z)
ylag1 = Z(:,1);
ylag2 = Z(:,2);
dydt = [ylag1(1);
ylag1(1)+ylag2(2);
y(2)];
end
Here In the call to ddex1de,
‘t’ is a scalar indicates the current ‘t’ in the equation
‘y’ is a column vector approximates y(t)
‘Z’ is a column vector approximates y(t – αj) for delay αj= lags(J).
In the following example Solution history is [1;1;1]. So Approximate values of lag ie ‘Z’ will be a matrix of size 3x2.In ‘Z’ Row defines the number of equations and column defines Number of lags.
4. Solve using ‘dde23’.
tspan = [0 5];
sol = dde23(@ddefun, lags, @history, tspan);

alpedhuez on 4 May 2020
Thank you.
1. Would you write down the content of this 3*2 matrix Z?
2. Another question:
function dydt = ddex1de(t,y,Z)
ylag1 = Z(:,1);
ylag2 = Z(:,2);
dydt = [ylag1(1);
ylag1(1)+ylag2(2);
y(2)];
end
I would think intuitively we define ddex1de by dydt so it should be like
function ddex1de = dydt(t,y,Z)
ylag1 = Z(:,1);
ylag2 = Z(:,2);
dydt = [ylag1(1);
ylag1(1)+ylag2(2);
y(2)];
end
Woud you explain why this is not the case?
Guru Mohanty on 4 May 2020
As it is a System of 3 Equations, 'y', 'dydt', 'ylag1' 'ylag2' have dimension 3x1. There are two constant Delay present, So The dimention of 'Z' is 3x2.